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    Essential fluid mechanics Essential fluid mechanics Document Transcript

    • FLUID MECHANICS {for energy conversion} Keith Vaugh BEng (AERO) MEng SME ASME
    • OBJECTIVES Identify the unique vocabulary associated with fluid mechanics with a focus on energy conservation } Explain the physical properties of fluids and derive the conservation laws of mass and energy for an ideal fluid (i.e. ignoring the viscous effects) Develop a comprehensive understanding of the effect of viscosity on the motion of a fluid flowing around an immersed body Determine the forces acting on an immersed body arising from the flow of fluid around itTo appreciate energy conversion such as hydro, wave, tidal and wind power a detailedknowledge of fluid mechanics is essential.During the course of this lecture, a brief summary of the basic physical properties of fluidsis provided and the conservation laws of mass and energy for an ideal (or inviscid) fluid arederived.The application of the conservation laws to situations of practical interest are also exploredto illustrate how useful information about the flow can be derived.Finally, the effect of viscosity on the motion of a fluid around an immersed body (such as aturbine blade) and how the flow determines the forces acting on the body of interest.
    • BULK PROPERTIES OF FLUIDS Density (ρ) - Mass per unit volume ⎛ ⎜ ⎝ m⎞ ρ= ⎟ V⎠ } Pressure (P) - Force per unit area in a fluid Viscosity - Force per unit area due to internal frictionDensity - Mass per unit volume of a fluid. For the purposes of this module,Density is assumed to be constant i.e. that the flow of the fluid isincompressible (small variations in pressure arising from fluid motion incomparison to atmospheric pressure).Pressure - Force per unit area, and acts in the normal direction to the surface ofa body in a fluid. Its units are the pascal or N-m^2Viscosity - Force per unit area due to internal friction in a fluid arising from therelative motion between neighbouring elements in a fluid. Viscous forces act inthe tangential direction to the surface of a body immersed in a flow. (consider adeck of cards.
    • STREAMLINES & STREAM-TUBES }A useful concept to visualize a velocity field is to imagine a set of streamlinespara;;e; to the direction of motion at all points in the fluid. Any element of massin the fluid flows along a notational stream-tube bounded by neighbouringstreamlines. In practice, streamlines can be visualised by injecting smallparticles into the fluid e.g. smoke can be used in wind tunnels to visualise fluidflow around objects.
    • MASS CONTINUITY Stream tube streamlines } velocity (u) Fluid stream ρuA = constAlso known as conservation of mass.Consider flow along a stream tube in a steady velocity field. The direction offlow is parallel to the boundries and at any point within the stream tube, thespeed of the fluid (u) and the cross-sectional area (A). Therefore the fluid isconfined to the stream tube and the mass flow per second is constant. Therefore:Density * Velocity * Cross-sectional Area = ConstantTherefore the speed of fluid is inversely proportional to the cross-sectional areaof the stream tube.
    • EXAMPLE 1An incompressible ideal fluid flows at a speed of 2 ms-1through a 1.2 m2 sectional area in which a constriction of }0.25 m2 sectional area has been inserted. What is the speedof the fluid inside the constriction?Putting ρ1 = ρ2, we have u1A1 = u2A2 A1 ⎛ 1.2 ⎞ u2 = u1 = (2 m s )⎜ ⎟ = 9.6 m s A2 ⎝ 0.25 ⎠
    • ENERGY OF A MOVING LIQUID The potential energy (PE) of liquid per unit mass ⎛ p⎞ } PE = g ⎜ z + ρg ⎟ p ⎝ ⎠ ρg and kinetic energy (KE) of liquid per unit mass u2 z KE = 2 total energy of liquid per unit mass ⎛ p u2 ⎞ Etotal = g⎜ z + + ⎝ ρ g 2g ⎟ ⎠Consider a liquid at a pressure p, moving with a velocity (u) at a height Z abovedatum levelIf a tube were inserted in the top of the pipe, the liquid would rise up the tube adistance of p/ρg and this is equivalent to an additional height of liquid relativeto datum level.Etotal equation represents the specific energy of the liquid. Each of the quantitiesin the brackets have the units of length and are termed heads, i.e Z is referred toas the potential head of the liquid, p/ρg as the pressure head and V2/2g as thevelocity head.
    • ENERGY CONSERVATION IN AN IDEAL FLUID In many practical situations, viscous forces are much smaller that those due to gravity and pressure gradients over large regions of the flow field. We can ignore viscosity to a good approximation and } derive an equation for energy conservation in a fluid known as Bernoulli’s equation. For steady flow, Bernoulli’s equation is of the form; p 1 + gz + u 2 = const ρ 2 For a stationary fluid, u = 0 p + gz = const ρFor a stationary fluid, u = 0 everywhere in the fluid therefore this equationreduces towhich is the equation for hydrostatic pressure. It shows that the fluid at a givendepth is all the same at the same pressure.
    • BERNOULLI’s EQUATION FOR STEADY FLOWConsider the steady flow of an ideal fluid in the control volume shown.The height (z), cross sectional area (A), speed (u) and pressure are denoted.The increase in gravitational potential energy of a mass δm of fluid between z1and z2 is δmg(z2-z1)In a short time interval (δt) the mass of fluid entering the control volume at P1is δm=ρu1A1δt and the mass exiting at P2 is δm=ρu2A2δt
    • By energy conservation, this is equal to the increase in Potential Energy plus the increase in Kinetic Energy, therefore; p1 A1u1δ t + p2 A2u2δ t = Work done in pushing elemental mass δm δ mg ( z2 − z1 ) + δ m ( u2 − u12 ) 1 2 a small distance δs 2 δ s = uδ t Putting δ m = ρu1 A1δ t = ρu2 A2δ t and tidying δW1 = p1 A1δ s1 = p1 A1u1δ t p1 1 p 1 2 + gz1 + u12 = 2 + gz2 + u2 Similarly at P2 ρ 2 ρ 2 δW2 = p2 A2u2δ t Finally, since P1 and P2 are arbitrary The net work done is p 1 δW2 + δW2 = p1 A1u1δ t + p2 A2u2δ t + gz + u 2 = const ρ 2In order for the fluid to enter the control volume it has to do work to overcomethe pressure p1 exerted by the fluid. The work done in pushing the elementalmass δm a small distance δs = uδtat P1 is δW1= Pressure * Area * Distance moved = Pressure * Area * Velocity *Change in time. Remember F = P*ASimilarly, the work done in pushing the elemental mass out of the controlvolume at P2 is δW2=-p2A2u2δt.The net work done is δW1+δW2
    • No losses between stations and u2 p1 u12 2 p2 u2 p2 z1 + + = z2 + + (1) ρg 2 ρg 2 u1 p1 Losses between stations and z2 z1 p1 u12 2 p2 u2 z1 + + = z2 + + +h (2) ρg 2 ρg 2 Energy additions or extractions p1 u12 win 2 p2 u2 wout z1 + + + = z2 + + + +h (3) ρg 2 g ρg 2 gConsider stations and of the inclined pipe illustrated. If there are nolosses between these two sections, and no energy changes resulting from heattransfer of work, then the total energy will remain constant, therefore stations and can be set equal to one another, equ (1)If however losses do occur between the two stations, then equation equ (2)When energy is added to the fluid by a device such as a pump, or when energyis extracted by a turbine, these need also be accounted for equ (3). w is thespecific work in J/kgNOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation
    • QUESTION’s(a) The atmospheric pressure on a surface is 105 Nm-2. Assuming the water is stationary, what is the pressure at a depth of 10 m ? } (assume ρwater = 103 kgm-3 & g = 9.81 ms-2)(b) What is the significance of Bernoulli’s equation?(c) Assuming the pressure of stationary air is 105 Nm-2, calculate the percentage change due to a wind of 20ms-1. (assume ρair ≈ 1.2 kgm-3)
    • EXAMPLE 2 }A Pitot tube is a device for measuring the velocity in a fluid.Essentially, it consists of two tubes, (a) and (b). Each tubehas one end open to the fluid and one end connected to apressure gauge. Tube (a) has the open end facing the flowand tube (b) has the open end normal to the flow. In the casewhen the gauge (a) reads p = po+pU2 and gauge (b) readsp = po, derive an expression for the velocity of the fluid interms of the difference in pressure between the gauges.
    • Bernoulli’s equation p 1 + gz + u 2 = const ρ 2 p0 1 2 ps Applying bernoulli’s equation + U = ρ 2 ρ ⎡ 2 ( ps − p0 ) ⎤ 1 2 Rearranging U=⎢ ⎥ ⎣ ρ ⎦Consider the fluid flowing along the stream-line AB. In case (a), the fluid slowsdown as it approaches the stagnation point B. Putting u = U, p = po at A and u= 0, p = ps at B and applying bernoulli’s equation we getIn example (and the previous question), ps is measured by tube (a) and p0measured by tube (b).NOTE - ps is larger than p0 by an amount ps - p0 = ρU2. The quantities ρU2,p0 and ps = p0 + ρU2 are called the dynamic pressure, static pressure and totalpressure respectively.
    • VELOCITY DISTRIBUTION & FLOW RATE The velocity in a flow stream varies considerably over the cross- section. A pitot tube only measures at one particular point, } therefore, in order to determine the velocity distribution over the entire cross section a number of readings are required.The velocity profile is obtained by traversing the pitot tube along the line AA.The cross-section is divided into convenient areas, a1, a2, a3, etc... and thevelocity at the centre of each area is determined
    • The flow rate can be determine; V = a1u1 + a2u2 + a3u3 + ... = ∑ au If the total cross-sectional area if the section is A; the mean velocity, U =  V = ∑ au A AThe velocity profile for a circular pipe is the same across any diameter  ΔV = ur × Δa where ur is the velocity at radius r V = ∑ ur × Δa  and V= ∑u r × Δa πr2The velocity at the boundary of any duct or pipe is zero.
    • EXAMPLE 3In a Venturi meter an ideal fluid flows with a volume flowrate and a pressure p1 through a horizontal pipe of cross }sectional area A1. A constriction of cross-sectional area A2 isinserted in the pipe and the pressure is p2 inside theconstriction. Derive an expression for the volume flow ratein terms of p1, p2, A1 and A2.
    • From Bernoulli’s equation p1 1 2 p2 1 2 + u = + u ρ 2 1 ρ 2 2Also by mass continuity A1 u2 = u1 A2Eliminating u2 we obtain the volume flow rate as ⎞ ⎡ 2 ( p1 − p2 ) ⎤ 1 2 ( V = A1u1 = ⎛ A1 A2 A12 − A ) 1  2 − 2 ⎝ 2 ⎠⎢ ρ ⎥ ⎣ ⎦
    • Applying Bernoulli’s equation to stations and h p1 u12 2 p2 u2 z1 + + = z2 + + ρg 2 ρg 2 Given that meter is horizontal, u1 u2 this equation reduces to: p1 u12 p2 u2 2 + = + ρg 2 ρg 2 A1 Mass continuity A1u1 = A2u2 ⇒ u2 = u1 A2 2 p1 p2 u12 ⎛ A1 ⎞ u12 u12 ⎛ A12 − A2 ⎞ 2 ∴ − =h= − = ρg ρg 2g ⎜ A2 ⎟ ⎝ ⎠ 2g 2g ⎜ A2 ⎟ ⎝ 2 ⎠A venturi meter is a device which is used to measure the rate of flow through apipe.NOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation
    • A2 ∴u1 = ( 2gh ) ( A12 − A2 2 ) A1 A2V = A1u1 = ( 2gh ) = k h since A1 and A2 are constants (A 2 1 −A 2 2 )The pressure at the throat is slightly lower than the the theoretical due tofriction in the tapered pipe. As a consequence h becomes slightly largerand therefore resulting in volumetric flow rate which is too high. Tocompensate for this, a coefficient of discharge, cd is introduced.  V = cd k h
    • The pressure differential between the main and throat sections is generallymeasured with a mercury-and-water U-tube, therefore h = x ( S − 1)If a Venturi meter is inclined to the horizontal, i.e station has a height z1and station a height z2 then;  V= k (h − ( z 2 − z1 ) )Show that the manometer reading for an inclined Venturi is the same as fora horizontal Venturi for a given flow rate.
    • ORIFICE PLATE Applying Bernoulli’s equation to stations and p1 u12 p2 u2 2 } + = + ρg 2 ρg 2 Mass continuity A1 A1u1 = A2u2 ⇒ u2 = u1 A2 ⎞ ⎡ 2 ( p1 − p2 ) ⎤ 1 2 ( ∴ V = A1u1 = ⎛ A1 A2 A12 − A ) 1  2 − 2 =k h ⎝ 2 ⎠⎢ ρ ⎥ ⎣ ⎦An Orifice plate is another means of determining the fluid flow in a pipe andworks on the same principle of the Venturi meter.The position of the Vena contract can be difficult establish and therefore thearea A2. The constant k is determined experimentally when it will incorporatethe coefficient of discharge.If h is small so that ρ is approximately constant then the this equation can beused for compressible fluid flow.NOTE - V1 and V2 indicated in diagram are u1 and u2 in derivation
    • EXAMPLE 4In an engine test 0.04 kg/s of air flows through a 50 mmdiameter pipe, into which is fitted a 40 mm diameter Orifice }plate. The density of air is 1.2 kg/m3 and the coefficient ofdischarge for the orifice is 0.63. The pressure drop acrossthe orifice plate is measured by a U-tube manometer.Calculate the manometer reading.
    •  Mass flow rate - m = ρV ⇒∴ V=0.033 m s  3  V= k h ⎛ π × 0.05 2 ⎞ ⎛ π × 0.04 2 ⎞ ⎜ ⎝ 4 ⎟⎜ ⎠⎝ 4 ⎟ ⎠  V= 1 2 × 9.81 × h ⎛ ⎛ π × 0.04 2 ⎞ 2 ⎛ π × 0.04 2 ⎞ 2 ⎞ 2 ⎜⎜ ⎝ 4 ⎟ −⎜ ⎠ ⎝ 4 ⎟ ⎟ ⎠ ⎠ ⎝  V = 0.007244 hThe actual discharge is 0.63 × 0.007244 h = 0.0333 ∴ h = 53.24m of airEquating pressures at Level XX in the U-tube 1.2g × 53.34 = 10 3 g × x ∴ x = 64 × 10 3 m
    • DYNAMICS OF A VISCOUS FLUID The motion of a viscous fluid is more complicated than that of an inviscid fluid. } The viscous shear force per unit area in the fluid is proportional to the velocity gradient y u ( y) = U F du u d = −µ =µ A dy d (0 ≤ y ≤ d) µ - coefficient of dynamic viscosityLet us consider a simple case of laminar flow between two parallel platesseparated by a small distance d.The upper plate moves at a constant velocity U while the lower plate remains atrest. At the plate fluid interface in both cases there is no velocity due to thestrong forces of attraction. Therefore the velocity profile in the fluid is given by
    • Typical flow regimes Laminar flow in a pipe Turbulent flow in a pipe ρUL UL Reynolds Number, Re = = µ vA viscous fluid can exhibit two different kinds of flow regimes, Laminar andTurbulentIn laminar flow, the fluid slides along distinct stream-tubes and tends to be quitestable, Turbulent flow is disorderly and unstableThe flow that exists in an y given situation depends on the ratio of the inertialforce to the viscous force. The magnitude of this ratio is given by Reynoldsnumber, where U is the velocity, L is the length and v = µ/ρ is the kinematicviscosity. Reynolds observed that flows at small Re where laminar while flow athigh Re contained regions of turbulence
    • Flow around a cylinder Flow around a cylinder for a for an inviscid fluid viscous fluidFor the inviscid flow the velocity fields in the upstream and downstreamregions are symmetrical, therefore the corresponding pressure distribution issymmetrical. It follows that the net force exerted by the fluid on the cylinder iszero. This is an example of d’Alembert’s paradoxFor a body immersed in a viscious fluid, the component of velocity tangential tothe surface of the fluid is zero at all points. At large Re numbers (Re>>1), theviscous force is negligible in the bulk of the fluid but us very significant in aviscous boundary layer close to the surface of the body. Rotational componentsof flow know as vorticity are generated within the boundary layer. As a certainpoint, the seperation point, the boundary layer becomes detached from thesurface and vorticity are discharged into the body of the fluid. The vorticity aretransported downstream side of the cylinder in the wake. Therefore the pressuredistributions on the upstream and downstream sides of the cylinder are notsymmetrical in the case of a viscous fluid. As a result, the cylinder experiencesa net force in the direction of motion known as drag. In the case of a spinningcylinder, a force called life arises at right angles to the direction of flow.
    • Lift and Drag Lift = 1 2 C L ρU 2 A Drag = 1 2 C D ρU 2 ACL and CD are dimensionless constants know as the coefficient of Lift and thecoefficient of Drag respectively.Lift and drag can be changed by altering the shape of a wing i.e. ailerons. Forsmall angles of attack, the pressure distribution ont he upper surface of anaerofoil is significantly lower than that on the lower surface which results is in anet lift force.
    • Variation of the lift and drag coefficients CL and CD with angle of attackCL and CD are dimensionless constants know as the coefficient of Lift and thecoefficient of Drag respectively.Lift and drag can be changed by altering the shape of a wing i.e. ailerons. Forsmall angles of attack, the pressure distribution on the upper surface of anaerofoil is significantly lower than that on the lower surface which results is in anet lift force.
    • Bulk properties of fluidsStreamlines & stream-tubesMass continuityEnergy conservation in an ideal fluidBernoulli’s equation for steady flow  Applied example’s (pitot tube, Venturi meter & questionsDynamics of a viscous fluid  Flow regimes  Laminar and turbulent flow  Vortices  Lift & Drag
    • Andrews, J., Jelley, N., (2007) Energy science: principles, technologies andimpacts, Oxford University PressBacon, D., Stephens, R. (1990) Mechanical Technology, second edition,Butterworth HeinemannBoyle, G. (2004) Renewable Energy: Power for a sustainable future, secondedition, Oxford University PressÇengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluidsciences, Third edition, McGraw HillTurns, S. (2006) Thermal fluid sciences: An integrated approach, CambridgeUniversity PressTwidell, J. and Weir, T. (2006) Renewable energy resources, second edition,Oxon: Taylor and FrancisIllustrations taken from Energy science: principles, technologies and impacts & Fundamentals of thermal fluid science